A uniform current carrying ring of mass \(m\) and radius \(R\) is connected by a massless string as shown. A uniform Magnetic field \(B_0\) exist in the region to keep the ring in horizontal position, then the current in the ring is

\((a)\) \(\frac{mg}{\pi R B_0}\)

\((b)\) \(\frac{mg}{ R B_0}\)

\((c)\) \(\frac{mg}{3 \pi R B_0}\)

\((d)\) \(\frac{mg}{\pi R^2 B_0}\)

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TopNewestThe tension of the string is \(T\), and the weight force on the loop is \(mg\). Because of the symmetry of the loop, the force caused by the magnetic field on the left half of the loop is equal and opposite to the force caused by the magnetic field on the right half. So the two net forces acting on the loop are \(T\) and \(mg\). The loop is at rest, so they must be equal.

\(T=mg\).

The torque on the loop due to the string is \(RT\), and the torque on the loop due to the magnetic field is \(B_0 A I\), where \(A\) is the area of the loop and \(I\) is the current. The area of the loop is \(A=\pi R^2\), so the torque due to the magnetic field is \(\pi R^2 B_0 I\). The loop is at rest, so the two torques must be equal.

\(RT=\pi R^2 B_0 I\)

\(Rmg=\pi R^2 B_0 I\)

\(I=\frac{mg}{\pi R B_0} \quad (a) \). – Ricky Escobar · 4 years ago

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T=mg (Force Balance).

TR=i\(\pi\)\(R^{2}\)\(B_{0}\) (Torque Balance)

Solve the two to get answer as (a). – Nishant Sharma · 4 years ago

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Is ur Age 14 Mr. Advitiya B. ? – Vamsi Krishna Appili · 4 years ago

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– Vamsi Krishna Appili · 4 years ago

Did u Qualify in NTSE ?Log in to reply

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I am curious ...becoz... My age person .in my Country.. . went to high Levels in challenges in Less TIME ... U joined from March 2013 right – Vamsi Krishna Appili · 4 years ago

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c) T + (2

PIRBI) = Mg ....... BALANCING THE FORCES TR=(PIRRBi) ..................BALANCING TORQUE SOLVE TO GET i :) – Vasanth Balakrishnan · 4 years agoLog in to reply